a² + 2ab + b² - ac - bc 

= ( a² + 2ab + b² ) - ac - bc 

= ( a + b ) ² - c ( a - b ) 

\(a^2+2ab+b^2-ac-bc=\left(a+b\right)^2-c\left(a+b\right)\)

                                                           \(=\left(a+b\right)\left(a+b-c\right)\)

Chúc bạn học tốt.

\(a^2+b^2+c^2-ab-bc-ca=0\)

\(\Leftrightarrow\)\(2\left(a^2+b^2+c^2-ab-bc-ca\right)=0\)

\(\Leftrightarrow\)\(\left(a^2-2ab+b^2\right)+\left(b^2-2bc+c^2\right)+\left(c^2-2ac+a^2\right)=0\)

\(\Leftrightarrow\)\(\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2=0\)

\(\Leftrightarrow\)\(\hept{\begin{cases}a-b=0\\b-c=0\\c-a=0\end{cases}}\)\(\Leftrightarrow\)\(\hept{\begin{cases}a=b\\b=c\\c=a\end{cases}}\)\(\Leftrightarrow\)\(a=b=c\)

=> đpcm

a)Ta có:a^2+b^2+c^2-ab-bc-ca=0

<=>2(a^2+b^2+c^2-ab-bc-ca)=2.0

<=>2a^2+2b^2+2c^2-2ab-2bc-2ca=0

<=>(a^2-2ab+b^2)+(b^2-2bc+c^2)+(c^2-2ca+a^2)=0

<=>(a-b)^2+(b-c)^2+(c-a)^2=0

Vì (a-b)^2>=0 với mọi a,b

    (b-c)^2>=0 với mọi b,c

    (c-a)^2>=0 với mọi c,a

Do đó:(a-b)^2+(b-c)^2+(c-a)^2>=0

Dấu "=" xảy ra <=>a-b=0<=>a=b

                            b-c=0<=>b=c

                            c-a=0<=>c=a

hay a=b=c(đpcm)

Ta có :(5a-3b+8c)(5a-3b-8c)=((5a-3b)+8c)((5a-3b)-8c)=(5a-3b)^2-(8c)^2=25a^2-30ab+9b^2-64c^2

=25a^2-30ab+9b^2-16.4c^2=25a^2-30ab+9b^2-16.4c^2(*)

Thay a^2-b^2=4c^2 vào(*)=>25a^2-30ab+9b^2-16(a^2-b^2)=25a^2-30ab+9b^2-16a^2+16b^2

=9a^2-30ab+25b^2=(3a)^2-3a.2.5b+(5b)^2=(3a-5b)^2

=>đpcm

Somebody help me

\(5y^3-10xy^2+5yx^2-20y=5y\left(y^2-2xy+x^2-4\right)\)

                                                              \(=5y\left[\left(x+y\right)^2-4\right]\)

                                                               \(=5y\left(x+y-2\right)\left(x+y+2\right)\)

Chúc bạn học tốt.

a)  \(P=4x^2-4x-1=\left(4x^2-4x+1\right)-2=\left(2x-1\right)^2-2\ge-2\)

Vậy Min P = - 2  khi  x = 1/2

b) \(P=3x^2-12x-3=3\left(x^2-4x+4\right)-15=3\left(x-2\right)^2-15\ge-15\)

Vậy Min P = -15  khi  x = 2

c)  \(P=\left(x+1\right)\left(2x-1\right)=2x^2+x-1=2\left(x+\frac{1}{4}\right)^2-\frac{9}{8}\ge-\frac{9}{8}\)

Vậy Min P = -9/8  khi  x = -1/4

\(\left(2x^m+7y^n\right)^2=4x^{2m}+28x^my^n+49y^{2n}\)

\(\left[\left(x-3\right)-z\right]^2=\left(x-3\right)^2-2\left(x-3\right)z+z^2=x^2-6x+9-2xz+6z+z^2\)

\(\left(4a^2-3b^2\right)\left(3b^2+4a^2\right)=\left(4a^2\right)^2-\left(3b^2\right)^2=16a^4-9b^4\)

Tham khảo nhé~

a)  \(f\left(x\right)=x^3+2x^2+x+2=0\)

\(\Leftrightarrow\)\(\left(x+2\right)\left(x^2+1\right)=0\)

\(\Leftrightarrow\)\(x+2=0\)

\(\Leftrightarrow\)\(x=-2\)

Vậy...

b)  \(g\left(x\right)=\left(x-1\right)x^2+\left(x-1\right)x+4x+4=0\)

\(\Leftrightarrow\)\(x\left(x-1\right)\left(x+1\right)+4\left(x+1\right)=0\)

\(\Leftrightarrow\)\(\left(x+1\right)\left(x^2-x+4\right)=0\)

\(\Leftrightarrow\)\(x+1=0\)

\(\Leftrightarrow\)\(x=-1\)

Vậy...

c)   \(h\left(x\right)=x^4-x^3+5x^2-5x=0\)

\(\Leftrightarrow\)\(x\left(x-1\right)\left(x^2+5\right)=0\)

\(\Leftrightarrow\)\(\orbr{\begin{cases}x=0\\x=1\end{cases}}\)

Vậy...